The rate constant of a reaction with nth order will have units:

Options

A. `litre^(1-n) mol^(n-1) sec^-1`
B. `mol^(1-n) litre^(1-n) sec^-1`
C. `mol^(1-n) litre^(n-1) sec^-1`
D. None of these

Answer

To determine the units of the rate constant (k) for a reaction of nth order, we can follow these steps:

### Step 1: Understand the Rate Law
The rate of a reaction can be expressed using the rate law:
\[ \text{Rate} = k [A]^n \]
where:
- Rate is the change in concentration over time (usually in mol/L/s),
- \( k \) is the rate constant,
- \([A]\) is the concentration of the reactant (in mol/L),
- \( n \) is the order of the reaction.

### Step 2: Rearranging the Rate Law
From the rate law, we can express the rate constant \( k \):
\[ k = \frac{\text{Rate}}{[A]^n} \]

### Step 3: Substitute Units
Now, we substitute the units into the equation:
- The unit of Rate is \( \text{mol/L/s} \),
- The unit of concentration \([A]\) is \( \text{mol/L} \).

Thus, substituting these into the equation gives:
\[ k = \frac{\text{mol/L/s}}{(\text{mol/L})^n} \]

### Step 4: Simplifying the Units
Now, simplify the right side:
\[ k = \frac{\text{mol/L/s}}{\text{(mol}^n/\text{L}^n)} = \frac{\text{mol}}{\text{L} \cdot \text{s}} \cdot \frac{\text{L}^n}{\text{mol}^n} \]

This can be simplified to:
\[ k = \frac{\text{mol}^{1-n}}{\text{L}^{1-n} \cdot \text{s}} \]

### Step 5: Final Units of Rate Constant
Thus, the final units of the rate constant \( k \) for a reaction of nth order is:
\[ k = \text{mol}^{1-n} \cdot \text{L}^{n-1} \cdot \text{s}^{-1} \]

### Conclusion
The correct answer for the units of the rate constant of a reaction with nth order is:
\[ \text{mol}^{1-n} \cdot \text{L}^{n-1} \cdot \text{s}^{-1} \]